Find all positive integers $a$ and $b$ such that $(1 + a)(8 + b)(a + b) = 27ab$.












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Here's the problem I'm having difficulties with:




Find all positive integers $a$ and $b$ such that $$(1 + a)(8 + b)(a + b) = 27ab,.$$




Does anyone have an idea how to do this? Any detailed solution is welcome! :)






number-theory elementary-number-theory contest-math diophantine-equations problem-solving







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  • 3




    $begingroup$
    What kind of numbers are $a$ and $b$? Rationals? Integers? Nonnegative integers? Positive integers?
    $endgroup$
    – Batominovski
    Dec 8 '18 at 18:03












  • $begingroup$
    Forgot to mention. I've edited the question. Thanks
    $endgroup$
    – Wolf M.
    Dec 8 '18 at 18:09






  • 2




    $begingroup$
    What are your thoughts? Put your work there and upload it.
    $endgroup$
    – jayant98
    Dec 8 '18 at 18:17






  • 2




    $begingroup$
    From a quick glance, we have $(a+1)(a+b)>ab$, so $$b+8=frac{27ab}{(a+1)(a+b)}<27,.$$ Thus, $bin{1,2,ldots,18}$. This shouldn't be too hard now. If you run out of ideas, you can still check all the $18$ cases ($b=1,2,ldots,18$), which is probably not too much work. There will be $18$ quadratic equations in $a$.
    $endgroup$
    – Batominovski
    Dec 8 '18 at 18:28








  • 1




    $begingroup$
    From $LHS$ one knows $b$ divides $1+a$ or $8a$; and $a$ divides $8+b$ or $b$.
    $endgroup$
    – AdditIdent
    Dec 8 '18 at 18:32
















1












$begingroup$


Here's the problem I'm having difficulties with:




Find all positive integers $a$ and $b$ such that $$(1 + a)(8 + b)(a + b) = 27ab,.$$




Does anyone have an idea how to do this? Any detailed solution is welcome! :)






number-theory elementary-number-theory contest-math diophantine-equations problem-solving







share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    What kind of numbers are $a$ and $b$? Rationals? Integers? Nonnegative integers? Positive integers?
    $endgroup$
    – Batominovski
    Dec 8 '18 at 18:03












  • $begingroup$
    Forgot to mention. I've edited the question. Thanks
    $endgroup$
    – Wolf M.
    Dec 8 '18 at 18:09






  • 2




    $begingroup$
    What are your thoughts? Put your work there and upload it.
    $endgroup$
    – jayant98
    Dec 8 '18 at 18:17






  • 2




    $begingroup$
    From a quick glance, we have $(a+1)(a+b)>ab$, so $$b+8=frac{27ab}{(a+1)(a+b)}<27,.$$ Thus, $bin{1,2,ldots,18}$. This shouldn't be too hard now. If you run out of ideas, you can still check all the $18$ cases ($b=1,2,ldots,18$), which is probably not too much work. There will be $18$ quadratic equations in $a$.
    $endgroup$
    – Batominovski
    Dec 8 '18 at 18:28








  • 1




    $begingroup$
    From $LHS$ one knows $b$ divides $1+a$ or $8a$; and $a$ divides $8+b$ or $b$.
    $endgroup$
    – AdditIdent
    Dec 8 '18 at 18:32














1












1








1


1



$begingroup$


Here's the problem I'm having difficulties with:




Find all positive integers $a$ and $b$ such that $$(1 + a)(8 + b)(a + b) = 27ab,.$$




Does anyone have an idea how to do this? Any detailed solution is welcome! :)






number-theory elementary-number-theory contest-math diophantine-equations problem-solving







share|cite|improve this question











$endgroup$




Here's the problem I'm having difficulties with:




Find all positive integers $a$ and $b$ such that $$(1 + a)(8 + b)(a + b) = 27ab,.$$




Does anyone have an idea how to do this? Any detailed solution is welcome! :)






number-theory elementary-number-theory contest-math diophantine-equations problem-solving




number-theory elementary-number-theory contest-math diophantine-equations problem-solving






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 8 '18 at 18:26









Batominovski

1




1










asked Dec 8 '18 at 18:01









Wolf M.Wolf M.

1097




1097








  • 3




    $begingroup$
    What kind of numbers are $a$ and $b$? Rationals? Integers? Nonnegative integers? Positive integers?
    $endgroup$
    – Batominovski
    Dec 8 '18 at 18:03












  • $begingroup$
    Forgot to mention. I've edited the question. Thanks
    $endgroup$
    – Wolf M.
    Dec 8 '18 at 18:09






  • 2




    $begingroup$
    What are your thoughts? Put your work there and upload it.
    $endgroup$
    – jayant98
    Dec 8 '18 at 18:17






  • 2




    $begingroup$
    From a quick glance, we have $(a+1)(a+b)>ab$, so $$b+8=frac{27ab}{(a+1)(a+b)}<27,.$$ Thus, $bin{1,2,ldots,18}$. This shouldn't be too hard now. If you run out of ideas, you can still check all the $18$ cases ($b=1,2,ldots,18$), which is probably not too much work. There will be $18$ quadratic equations in $a$.
    $endgroup$
    – Batominovski
    Dec 8 '18 at 18:28








  • 1




    $begingroup$
    From $LHS$ one knows $b$ divides $1+a$ or $8a$; and $a$ divides $8+b$ or $b$.
    $endgroup$
    – AdditIdent
    Dec 8 '18 at 18:32














  • 3




    $begingroup$
    What kind of numbers are $a$ and $b$? Rationals? Integers? Nonnegative integers? Positive integers?
    $endgroup$
    – Batominovski
    Dec 8 '18 at 18:03












  • $begingroup$
    Forgot to mention. I've edited the question. Thanks
    $endgroup$
    – Wolf M.
    Dec 8 '18 at 18:09






  • 2




    $begingroup$
    What are your thoughts? Put your work there and upload it.
    $endgroup$
    – jayant98
    Dec 8 '18 at 18:17






  • 2




    $begingroup$
    From a quick glance, we have $(a+1)(a+b)>ab$, so $$b+8=frac{27ab}{(a+1)(a+b)}<27,.$$ Thus, $bin{1,2,ldots,18}$. This shouldn't be too hard now. If you run out of ideas, you can still check all the $18$ cases ($b=1,2,ldots,18$), which is probably not too much work. There will be $18$ quadratic equations in $a$.
    $endgroup$
    – Batominovski
    Dec 8 '18 at 18:28








  • 1




    $begingroup$
    From $LHS$ one knows $b$ divides $1+a$ or $8a$; and $a$ divides $8+b$ or $b$.
    $endgroup$
    – AdditIdent
    Dec 8 '18 at 18:32








3




3




$begingroup$
What kind of numbers are $a$ and $b$? Rationals? Integers? Nonnegative integers? Positive integers?
$endgroup$
– Batominovski
Dec 8 '18 at 18:03






$begingroup$
What kind of numbers are $a$ and $b$? Rationals? Integers? Nonnegative integers? Positive integers?
$endgroup$
– Batominovski
Dec 8 '18 at 18:03














$begingroup$
Forgot to mention. I've edited the question. Thanks
$endgroup$
– Wolf M.
Dec 8 '18 at 18:09




$begingroup$
Forgot to mention. I've edited the question. Thanks
$endgroup$
– Wolf M.
Dec 8 '18 at 18:09




2




2




$begingroup$
What are your thoughts? Put your work there and upload it.
$endgroup$
– jayant98
Dec 8 '18 at 18:17




$begingroup$
What are your thoughts? Put your work there and upload it.
$endgroup$
– jayant98
Dec 8 '18 at 18:17




2




2




$begingroup$
From a quick glance, we have $(a+1)(a+b)>ab$, so $$b+8=frac{27ab}{(a+1)(a+b)}<27,.$$ Thus, $bin{1,2,ldots,18}$. This shouldn't be too hard now. If you run out of ideas, you can still check all the $18$ cases ($b=1,2,ldots,18$), which is probably not too much work. There will be $18$ quadratic equations in $a$.
$endgroup$
– Batominovski
Dec 8 '18 at 18:28






$begingroup$
From a quick glance, we have $(a+1)(a+b)>ab$, so $$b+8=frac{27ab}{(a+1)(a+b)}<27,.$$ Thus, $bin{1,2,ldots,18}$. This shouldn't be too hard now. If you run out of ideas, you can still check all the $18$ cases ($b=1,2,ldots,18$), which is probably not too much work. There will be $18$ quadratic equations in $a$.
$endgroup$
– Batominovski
Dec 8 '18 at 18:28






1




1




$begingroup$
From $LHS$ one knows $b$ divides $1+a$ or $8a$; and $a$ divides $8+b$ or $b$.
$endgroup$
– AdditIdent
Dec 8 '18 at 18:32




$begingroup$
From $LHS$ one knows $b$ divides $1+a$ or $8a$; and $a$ divides $8+b$ or $b$.
$endgroup$
– AdditIdent
Dec 8 '18 at 18:32










3 Answers
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Using Hölder's inequality,
$$27ab = (a+1)(8+b)(b+a) geqslant left(2sqrt[3]{ab}+sqrt[3]{ab} right)^3=27ab$$



Hence we are looking for the equality case for Hölder, which is when $a:8:b=1:b:a implies (a, b)=(2, 4)$.



In fact, this is the only solution among positive reals, not just positive integers.






share|cite|improve this answer











$endgroup$





















    4












    $begingroup$

    This is a supplementary solution, where I solve for all $(a,b)inmathbb{Z}timesmathbb{Z}$ such that $$(1+a)(8+b)(a+b)=27ab,.$$ From $(1+a)(8+b)(a+b)-27ab=0$, we have
    $$(8+b)a^2+big((8+b)(b+1)-27bbig)a+b(8+b)=0,.$$
    The discriminant of this quadratic polynomial with respect to $a$ is
    $$begin{align}big((8+b)(b+1)-27bbig)^2-4cdot(8+b)cdot b(8+b)&=b^4-40b^3+276b^2-544b+64\&=(b-4)^2(b^2-32b+4),.end{align}$$
    We require that $(b-4)^2(b^2-32b+4)$ be a perfect square. If $b=4$, then $$12(a-2)^2=12left(a^2-4a+4right)=0,,$$ so $a=2$. If $bneq 4$, then $$(b-16)^2-252=b^2-32b+4=c^2$$
    for some integer $c$. Thus, $$d^2-c^2=252,,$$ where $d:=b-16$.



    Since $4mid 252$ but $8nmid 252$, both $c$ and $d$ are even. Let $c:=2p$ and $d:=2q$, so that $$(q+p)(q-p)=q^2-p^2=frac{d^2-c^2}{4}=63tag{*},.$$
    Therefore, the possible values of $(q+p,q-p)$ are $$(-63,-1),,,,(-21,-3),,,,(-9,-7),,,,(-7,-9),,,,(-3,-21),,,,(-1,-63),,$$ $$(1,63),,,,(3,21),,,,(7,9),,,,(9,7),,,,(21,3),,text{ and }(63,1),.$$
    Thus, $b-16=d=2q=(q+p)+(q-p)$ takes the $6$ values
    $$-64,-24,-16,+16,+24,+64,.$$
    Ergo, $bin{-48,-8,0,32,40,80}$, resulting in the following solutions $(a,b)$:
    $$(80,-48),,,,(0,-8),,,,(-1,0),,,,(0,0),,,,(-5,32),(-16,40),,text{ and }(-55,80),,$$
    as well as the pair $(2,4)$ found earlier.



    Using (*), we can also find all rational solutions. By setting $r:=q+p$, the rational solutions $(a,b)neq (2,4)$ take the form
    $$left(-frac{(3+r)(7+r)}{21+r},frac{(7+r)(9+r)}{r}right)text{ for }rinmathbb{Q}setminus{0,-21},,tag{#}$$
    and
    $$left(-frac{(9+r)(21+r)}{r(3+r)},frac{(7+r)(9+r)}{r}right)text{ for }rinmathbb{Q}setminus{0,-3},.tag{@}$$
    By the way, I just realized that with the transformation $rmapstodfrac{63}{r}$, the two solutions (#) and (@) are identical. (The same parametrization also works if you want to solve for real solutions $(a,b)neq (2,4)$, or even complex solutions $(a,b)$, where $r:=-6pm3sqrt{3}text{i}$ gives rise to the pair $(a,b)=(2,4)$.)






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      $begingroup$

      This is not a complete solution, but it points the way to one, and its simplicity, I think, makes it worth mentioning.



      By expanding out the product and rearranging the results, we obtain the equivalent equation to solve,



      $${8+8aover b}+{8+bover a}=18-a-b$$



      Since the left hand side is positive, the right hand side limits the possibilities for $a$ and $b$ to a small enough set for brute force to take over.






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        3 Answers
        3






        active

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        3 Answers
        3






        active

        oldest

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        active

        oldest

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        active

        oldest

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        4












        $begingroup$

        Using Hölder's inequality,
        $$27ab = (a+1)(8+b)(b+a) geqslant left(2sqrt[3]{ab}+sqrt[3]{ab} right)^3=27ab$$



        Hence we are looking for the equality case for Hölder, which is when $a:8:b=1:b:a implies (a, b)=(2, 4)$.



        In fact, this is the only solution among positive reals, not just positive integers.






        share|cite|improve this answer











        $endgroup$


















          4












          $begingroup$

          Using Hölder's inequality,
          $$27ab = (a+1)(8+b)(b+a) geqslant left(2sqrt[3]{ab}+sqrt[3]{ab} right)^3=27ab$$



          Hence we are looking for the equality case for Hölder, which is when $a:8:b=1:b:a implies (a, b)=(2, 4)$.



          In fact, this is the only solution among positive reals, not just positive integers.






          share|cite|improve this answer











          $endgroup$
















            4












            4








            4





            $begingroup$

            Using Hölder's inequality,
            $$27ab = (a+1)(8+b)(b+a) geqslant left(2sqrt[3]{ab}+sqrt[3]{ab} right)^3=27ab$$



            Hence we are looking for the equality case for Hölder, which is when $a:8:b=1:b:a implies (a, b)=(2, 4)$.



            In fact, this is the only solution among positive reals, not just positive integers.






            share|cite|improve this answer











            $endgroup$



            Using Hölder's inequality,
            $$27ab = (a+1)(8+b)(b+a) geqslant left(2sqrt[3]{ab}+sqrt[3]{ab} right)^3=27ab$$



            Hence we are looking for the equality case for Hölder, which is when $a:8:b=1:b:a implies (a, b)=(2, 4)$.



            In fact, this is the only solution among positive reals, not just positive integers.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Dec 8 '18 at 18:58









            Théophile

            19.5k12946




            19.5k12946










            answered Dec 8 '18 at 18:41









            MacavityMacavity

            35.2k52554




            35.2k52554























                4












                $begingroup$

                This is a supplementary solution, where I solve for all $(a,b)inmathbb{Z}timesmathbb{Z}$ such that $$(1+a)(8+b)(a+b)=27ab,.$$ From $(1+a)(8+b)(a+b)-27ab=0$, we have
                $$(8+b)a^2+big((8+b)(b+1)-27bbig)a+b(8+b)=0,.$$
                The discriminant of this quadratic polynomial with respect to $a$ is
                $$begin{align}big((8+b)(b+1)-27bbig)^2-4cdot(8+b)cdot b(8+b)&=b^4-40b^3+276b^2-544b+64\&=(b-4)^2(b^2-32b+4),.end{align}$$
                We require that $(b-4)^2(b^2-32b+4)$ be a perfect square. If $b=4$, then $$12(a-2)^2=12left(a^2-4a+4right)=0,,$$ so $a=2$. If $bneq 4$, then $$(b-16)^2-252=b^2-32b+4=c^2$$
                for some integer $c$. Thus, $$d^2-c^2=252,,$$ where $d:=b-16$.



                Since $4mid 252$ but $8nmid 252$, both $c$ and $d$ are even. Let $c:=2p$ and $d:=2q$, so that $$(q+p)(q-p)=q^2-p^2=frac{d^2-c^2}{4}=63tag{*},.$$
                Therefore, the possible values of $(q+p,q-p)$ are $$(-63,-1),,,,(-21,-3),,,,(-9,-7),,,,(-7,-9),,,,(-3,-21),,,,(-1,-63),,$$ $$(1,63),,,,(3,21),,,,(7,9),,,,(9,7),,,,(21,3),,text{ and }(63,1),.$$
                Thus, $b-16=d=2q=(q+p)+(q-p)$ takes the $6$ values
                $$-64,-24,-16,+16,+24,+64,.$$
                Ergo, $bin{-48,-8,0,32,40,80}$, resulting in the following solutions $(a,b)$:
                $$(80,-48),,,,(0,-8),,,,(-1,0),,,,(0,0),,,,(-5,32),(-16,40),,text{ and }(-55,80),,$$
                as well as the pair $(2,4)$ found earlier.



                Using (*), we can also find all rational solutions. By setting $r:=q+p$, the rational solutions $(a,b)neq (2,4)$ take the form
                $$left(-frac{(3+r)(7+r)}{21+r},frac{(7+r)(9+r)}{r}right)text{ for }rinmathbb{Q}setminus{0,-21},,tag{#}$$
                and
                $$left(-frac{(9+r)(21+r)}{r(3+r)},frac{(7+r)(9+r)}{r}right)text{ for }rinmathbb{Q}setminus{0,-3},.tag{@}$$
                By the way, I just realized that with the transformation $rmapstodfrac{63}{r}$, the two solutions (#) and (@) are identical. (The same parametrization also works if you want to solve for real solutions $(a,b)neq (2,4)$, or even complex solutions $(a,b)$, where $r:=-6pm3sqrt{3}text{i}$ gives rise to the pair $(a,b)=(2,4)$.)






                share|cite|improve this answer











                $endgroup$


















                  4












                  $begingroup$

                  This is a supplementary solution, where I solve for all $(a,b)inmathbb{Z}timesmathbb{Z}$ such that $$(1+a)(8+b)(a+b)=27ab,.$$ From $(1+a)(8+b)(a+b)-27ab=0$, we have
                  $$(8+b)a^2+big((8+b)(b+1)-27bbig)a+b(8+b)=0,.$$
                  The discriminant of this quadratic polynomial with respect to $a$ is
                  $$begin{align}big((8+b)(b+1)-27bbig)^2-4cdot(8+b)cdot b(8+b)&=b^4-40b^3+276b^2-544b+64\&=(b-4)^2(b^2-32b+4),.end{align}$$
                  We require that $(b-4)^2(b^2-32b+4)$ be a perfect square. If $b=4$, then $$12(a-2)^2=12left(a^2-4a+4right)=0,,$$ so $a=2$. If $bneq 4$, then $$(b-16)^2-252=b^2-32b+4=c^2$$
                  for some integer $c$. Thus, $$d^2-c^2=252,,$$ where $d:=b-16$.



                  Since $4mid 252$ but $8nmid 252$, both $c$ and $d$ are even. Let $c:=2p$ and $d:=2q$, so that $$(q+p)(q-p)=q^2-p^2=frac{d^2-c^2}{4}=63tag{*},.$$
                  Therefore, the possible values of $(q+p,q-p)$ are $$(-63,-1),,,,(-21,-3),,,,(-9,-7),,,,(-7,-9),,,,(-3,-21),,,,(-1,-63),,$$ $$(1,63),,,,(3,21),,,,(7,9),,,,(9,7),,,,(21,3),,text{ and }(63,1),.$$
                  Thus, $b-16=d=2q=(q+p)+(q-p)$ takes the $6$ values
                  $$-64,-24,-16,+16,+24,+64,.$$
                  Ergo, $bin{-48,-8,0,32,40,80}$, resulting in the following solutions $(a,b)$:
                  $$(80,-48),,,,(0,-8),,,,(-1,0),,,,(0,0),,,,(-5,32),(-16,40),,text{ and }(-55,80),,$$
                  as well as the pair $(2,4)$ found earlier.



                  Using (*), we can also find all rational solutions. By setting $r:=q+p$, the rational solutions $(a,b)neq (2,4)$ take the form
                  $$left(-frac{(3+r)(7+r)}{21+r},frac{(7+r)(9+r)}{r}right)text{ for }rinmathbb{Q}setminus{0,-21},,tag{#}$$
                  and
                  $$left(-frac{(9+r)(21+r)}{r(3+r)},frac{(7+r)(9+r)}{r}right)text{ for }rinmathbb{Q}setminus{0,-3},.tag{@}$$
                  By the way, I just realized that with the transformation $rmapstodfrac{63}{r}$, the two solutions (#) and (@) are identical. (The same parametrization also works if you want to solve for real solutions $(a,b)neq (2,4)$, or even complex solutions $(a,b)$, where $r:=-6pm3sqrt{3}text{i}$ gives rise to the pair $(a,b)=(2,4)$.)






                  share|cite|improve this answer











                  $endgroup$
















                    4












                    4








                    4





                    $begingroup$

                    This is a supplementary solution, where I solve for all $(a,b)inmathbb{Z}timesmathbb{Z}$ such that $$(1+a)(8+b)(a+b)=27ab,.$$ From $(1+a)(8+b)(a+b)-27ab=0$, we have
                    $$(8+b)a^2+big((8+b)(b+1)-27bbig)a+b(8+b)=0,.$$
                    The discriminant of this quadratic polynomial with respect to $a$ is
                    $$begin{align}big((8+b)(b+1)-27bbig)^2-4cdot(8+b)cdot b(8+b)&=b^4-40b^3+276b^2-544b+64\&=(b-4)^2(b^2-32b+4),.end{align}$$
                    We require that $(b-4)^2(b^2-32b+4)$ be a perfect square. If $b=4$, then $$12(a-2)^2=12left(a^2-4a+4right)=0,,$$ so $a=2$. If $bneq 4$, then $$(b-16)^2-252=b^2-32b+4=c^2$$
                    for some integer $c$. Thus, $$d^2-c^2=252,,$$ where $d:=b-16$.



                    Since $4mid 252$ but $8nmid 252$, both $c$ and $d$ are even. Let $c:=2p$ and $d:=2q$, so that $$(q+p)(q-p)=q^2-p^2=frac{d^2-c^2}{4}=63tag{*},.$$
                    Therefore, the possible values of $(q+p,q-p)$ are $$(-63,-1),,,,(-21,-3),,,,(-9,-7),,,,(-7,-9),,,,(-3,-21),,,,(-1,-63),,$$ $$(1,63),,,,(3,21),,,,(7,9),,,,(9,7),,,,(21,3),,text{ and }(63,1),.$$
                    Thus, $b-16=d=2q=(q+p)+(q-p)$ takes the $6$ values
                    $$-64,-24,-16,+16,+24,+64,.$$
                    Ergo, $bin{-48,-8,0,32,40,80}$, resulting in the following solutions $(a,b)$:
                    $$(80,-48),,,,(0,-8),,,,(-1,0),,,,(0,0),,,,(-5,32),(-16,40),,text{ and }(-55,80),,$$
                    as well as the pair $(2,4)$ found earlier.



                    Using (*), we can also find all rational solutions. By setting $r:=q+p$, the rational solutions $(a,b)neq (2,4)$ take the form
                    $$left(-frac{(3+r)(7+r)}{21+r},frac{(7+r)(9+r)}{r}right)text{ for }rinmathbb{Q}setminus{0,-21},,tag{#}$$
                    and
                    $$left(-frac{(9+r)(21+r)}{r(3+r)},frac{(7+r)(9+r)}{r}right)text{ for }rinmathbb{Q}setminus{0,-3},.tag{@}$$
                    By the way, I just realized that with the transformation $rmapstodfrac{63}{r}$, the two solutions (#) and (@) are identical. (The same parametrization also works if you want to solve for real solutions $(a,b)neq (2,4)$, or even complex solutions $(a,b)$, where $r:=-6pm3sqrt{3}text{i}$ gives rise to the pair $(a,b)=(2,4)$.)






                    share|cite|improve this answer











                    $endgroup$



                    This is a supplementary solution, where I solve for all $(a,b)inmathbb{Z}timesmathbb{Z}$ such that $$(1+a)(8+b)(a+b)=27ab,.$$ From $(1+a)(8+b)(a+b)-27ab=0$, we have
                    $$(8+b)a^2+big((8+b)(b+1)-27bbig)a+b(8+b)=0,.$$
                    The discriminant of this quadratic polynomial with respect to $a$ is
                    $$begin{align}big((8+b)(b+1)-27bbig)^2-4cdot(8+b)cdot b(8+b)&=b^4-40b^3+276b^2-544b+64\&=(b-4)^2(b^2-32b+4),.end{align}$$
                    We require that $(b-4)^2(b^2-32b+4)$ be a perfect square. If $b=4$, then $$12(a-2)^2=12left(a^2-4a+4right)=0,,$$ so $a=2$. If $bneq 4$, then $$(b-16)^2-252=b^2-32b+4=c^2$$
                    for some integer $c$. Thus, $$d^2-c^2=252,,$$ where $d:=b-16$.



                    Since $4mid 252$ but $8nmid 252$, both $c$ and $d$ are even. Let $c:=2p$ and $d:=2q$, so that $$(q+p)(q-p)=q^2-p^2=frac{d^2-c^2}{4}=63tag{*},.$$
                    Therefore, the possible values of $(q+p,q-p)$ are $$(-63,-1),,,,(-21,-3),,,,(-9,-7),,,,(-7,-9),,,,(-3,-21),,,,(-1,-63),,$$ $$(1,63),,,,(3,21),,,,(7,9),,,,(9,7),,,,(21,3),,text{ and }(63,1),.$$
                    Thus, $b-16=d=2q=(q+p)+(q-p)$ takes the $6$ values
                    $$-64,-24,-16,+16,+24,+64,.$$
                    Ergo, $bin{-48,-8,0,32,40,80}$, resulting in the following solutions $(a,b)$:
                    $$(80,-48),,,,(0,-8),,,,(-1,0),,,,(0,0),,,,(-5,32),(-16,40),,text{ and }(-55,80),,$$
                    as well as the pair $(2,4)$ found earlier.



                    Using (*), we can also find all rational solutions. By setting $r:=q+p$, the rational solutions $(a,b)neq (2,4)$ take the form
                    $$left(-frac{(3+r)(7+r)}{21+r},frac{(7+r)(9+r)}{r}right)text{ for }rinmathbb{Q}setminus{0,-21},,tag{#}$$
                    and
                    $$left(-frac{(9+r)(21+r)}{r(3+r)},frac{(7+r)(9+r)}{r}right)text{ for }rinmathbb{Q}setminus{0,-3},.tag{@}$$
                    By the way, I just realized that with the transformation $rmapstodfrac{63}{r}$, the two solutions (#) and (@) are identical. (The same parametrization also works if you want to solve for real solutions $(a,b)neq (2,4)$, or even complex solutions $(a,b)$, where $r:=-6pm3sqrt{3}text{i}$ gives rise to the pair $(a,b)=(2,4)$.)







                    share|cite|improve this answer














                    share|cite|improve this answer



                    share|cite|improve this answer








                    edited Dec 8 '18 at 20:08

























                    answered Dec 8 '18 at 19:07









                    BatominovskiBatominovski

                    1




                    1























                        2












                        $begingroup$

                        This is not a complete solution, but it points the way to one, and its simplicity, I think, makes it worth mentioning.



                        By expanding out the product and rearranging the results, we obtain the equivalent equation to solve,



                        $${8+8aover b}+{8+bover a}=18-a-b$$



                        Since the left hand side is positive, the right hand side limits the possibilities for $a$ and $b$ to a small enough set for brute force to take over.






                        share|cite|improve this answer









                        $endgroup$


















                          2












                          $begingroup$

                          This is not a complete solution, but it points the way to one, and its simplicity, I think, makes it worth mentioning.



                          By expanding out the product and rearranging the results, we obtain the equivalent equation to solve,



                          $${8+8aover b}+{8+bover a}=18-a-b$$



                          Since the left hand side is positive, the right hand side limits the possibilities for $a$ and $b$ to a small enough set for brute force to take over.






                          share|cite|improve this answer









                          $endgroup$
















                            2












                            2








                            2





                            $begingroup$

                            This is not a complete solution, but it points the way to one, and its simplicity, I think, makes it worth mentioning.



                            By expanding out the product and rearranging the results, we obtain the equivalent equation to solve,



                            $${8+8aover b}+{8+bover a}=18-a-b$$



                            Since the left hand side is positive, the right hand side limits the possibilities for $a$ and $b$ to a small enough set for brute force to take over.






                            share|cite|improve this answer









                            $endgroup$



                            This is not a complete solution, but it points the way to one, and its simplicity, I think, makes it worth mentioning.



                            By expanding out the product and rearranging the results, we obtain the equivalent equation to solve,



                            $${8+8aover b}+{8+bover a}=18-a-b$$



                            Since the left hand side is positive, the right hand side limits the possibilities for $a$ and $b$ to a small enough set for brute force to take over.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Dec 8 '18 at 20:54









                            Barry CipraBarry Cipra

                            59.4k653125




                            59.4k653125






























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